question_answer

\[\Delta ABC\]is right angled triangle, where \[\angle ABC=90{}^\circ .\]If \[AC=2\sqrt{5}\]and \[AB-BC=2,\] then the value of \[{{\cos }^{2}}A-{{\cos }^{2}}C\]is                                                                         [SSC (CGL) Mains 2012]

A) \[\frac{1}{\sqrt{5}}\]                             

B) \[\sqrt{5}\]

C) \[\frac{1}{2}\]                          

D) \[\frac{3}{5}\]

Correct Answer: D

Solution :

Let \[BC=x,\]then \[AB=x+2\]

In \[\Delta ABC,\]

\[{{(x+2)}^{2}}+{{x}^{2}}={{(2\sqrt{5})}^{2}}\]

\[\Rightarrow \]\[{{x}^{2}}+4+4x+{{x}^{2}}=20\]

\[\Rightarrow \]\[2{{x}^{2}}+4x-16=0\]

\[\Rightarrow \]\[{{x}^{2}}+2x-8=0\]

\[\Rightarrow \]\[(x-2)(x+4)=0\]

\[\Rightarrow \]               \[x=2\]

So,       \[AB=4,\]\[BC=2\]

\[\therefore \]\[{{\cos }^{2}}A-{{\cos }^{2}}C={{\left( \frac{4}{2\sqrt{5}} \right)}^{2}}-{{\left( \frac{2}{2\sqrt{5}} \right)}^{2}}=\frac{3}{5}\]